Question

Find k for the following equation to have equal and real roots.

$k {x}^{2} + 4x + 1 = 0$
Pls explain​

Answer 1

On comparing the given equation with

$a {x}^{2} + bx + c$

we get a=k, b=4, c=1

If roots are real and equal then D=0

Therefore,

${b}^{2} - 4ac = 0$

On putting the values of a,b and c

${4}^{2} - 4 \times k \times 1 = 0$

=》16-4k=0

=》4k=16

=》k=4